Complex Numbers — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: Definition of $i$, real and imaginary parts, modulus and argument, polar form operations, Argand diagram geometric representations, and loci including circles and perpendicular bisectors.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trigonometry).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Are Complex Numbers?

Complex numbers extend the real number system to solve equations that have no real solutions, such as $x^2=-1$. For the A-Level Mathematics P3 syllabus, you will use complex numbers to solve polynomial equations, interpret geometric relationships, and simplify calculations involving periodic or rotational phenomena. They appear on almost every Paper 3 exam, usually as 5-8 mark questions that combine algebraic manipulation and geometric reasoning.

2. Definition of $i$, real and imaginary parts

The core unit of complex numbers is the imaginary unit $i$, defined such that $i^2=-1$. A general complex number $z$ is written in Cartesian form as: $$z=a+bi$$ where $a$ is the real part of $z$, written $\text{Re}(z)=a$, and $b$ is the imaginary part of $z$, written $\text{Im}(z)=b$. Note that the imaginary part is the real coefficient of $i$, not including $i$ itself.

Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal: if $a+bi=c+di$, then $a=c$ and $b=d$. This is the key rule for solving equations with unknown complex numbers.

Worked Example

Given $2x+5yi=8-15i$, find the values of real constants $x$ and $y$.

1. Equate real parts: $2x=8⟹x=4$
2. Equate imaginary parts: $5y=-15⟹y=-3$

Exam tip: Examiners frequently test this rule with hidden unknowns, so always split complex equations into real and imaginary components first before solving.

3. Modulus and argument; polar form $r(\cos \theta +i\sin \theta )$

Every complex number can also be described using its distance from the origin of the number plane, and its angle relative to the positive real axis.

• The modulus of $z=a+bi$, written $|z|$, is the non-negative distance from the origin to the point representing $z$: $$|z|=r=\sqrt{a^2+b^2}$$
• The principal argument of $z$, written $\arg (z)$, is the angle $\theta$ (in radians) between the positive real axis and the line connecting the origin to $z$, measured counterclockwise. The principal argument always lies in the range $-\pi <\theta \le \pi$, so you must adjust your calculated angle to match this range based on the quadrant $z$ lies in.

Combining these gives the polar form of $z$: $$z=r(\cos \theta +i\sin \theta )$$ This is sometimes abbreviated as $rcis\theta$, but you should always write the full trigonometric form in exams to avoid losing marks.

Worked Example

Write $z=-1-i\sqrt{3}$ in polar form.

1. Calculate modulus: $|z|=\sqrt{(-1{)}^2+(-\sqrt{3}{)}^2}=\sqrt{1+3}=2$
2. Calculate base angle: $\arctan \left(\frac{|\text{Im}(z)|}{|\text{Re}(z)|}\right)=\arctan (\sqrt{3})=\frac{\pi }{3}$
3. $z$ lies in the third quadrant, so principal argument: $\theta =-\pi +\frac{\pi }{3}=-\frac{2\pi }{3}$
4. Polar form: $z=2\left(\cos \left(-\frac{2\pi }{3}\right)+i\sin \left(-\frac{2\pi }{3}\right)\right)$

4. Multiplication and division in polar form

Polar form simplifies complex number multiplication and division significantly, by using trigonometric addition identities you learned in Pure 1. For two complex numbers $z_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)$:

1. Multiplication: Multiply the moduli, add the arguments: $$z_1{z}_2=r_1{r}_2\left(\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)\right)$$
2. Division: Divide the moduli, subtract the argument of the divisor: $$\frac{z_1}{z_2}=\frac{r_1}{r_2}\left(\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)\right)$$

Worked Example

Given $z_1=3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ and $z_2=2\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$, find $z_1{z}_2$ and $\frac{z_1}{z_2}$ in Cartesian form.

1. Product: $|z_1{z}_2|=3\ast 2=6$, $\arg (z_1{z}_2)=\frac{\pi }{4}+\frac{\pi }{6}=\frac{5\pi }{12}$ $$z_1{z}_2=6\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)\approx 1.55+5.80i$$
2. Quotient: $\left|\frac{z_1}{z_2}\right|=\frac{3}{2}$, $\arg \left(\frac{z_1}{z_2}\right)=\frac{\pi }{4}-\frac{\pi }{6}=\frac{\pi }{12}$ $$\frac{z_1}{z_2}=1.5\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)\approx 1.45+0.39i$$

Exam tip: Always use polar form for calculations involving 3 or more complex multiplications/divisions, or powers of complex numbers. It is far faster and reduces the risk of arithmetic errors compared to Cartesian form.

5. Argand diagram — geometric representations

An Argand diagram is a 2D coordinate plane where the x-axis is the real axis and the y-axis is the imaginary axis. Every complex number $z=a+bi$ is represented as either the point $(a,b)$ or the vector from the origin to $(a,b)$.

Key geometric interpretations:

• Adding two complex numbers is equivalent to vector addition (parallelogram rule)
• Subtracting $w$ from $z$ gives the vector from $w$ to $z$
• The distance between two points $z$ and $w$ is $|z-w|$
• Multiplying a complex number by $i$ rotates its vector 90° counterclockwise around the origin, with no change to its modulus

Worked Example

Plot $z=2+i$ and $w=-1+3i$ on an Argand diagram, and show that the distance between them is $\sqrt{13}$.

1. Plot $z$ at $(2,1)$ and $w$ at $(-1,3)$
2. Calculate $|z-w|=|(2+1)+(1-3)i|=|3-2i|=\sqrt{3^2+(-2{)}^2}=\sqrt{13}$, which matches the linear distance between the two points.

6. Loci on the Argand diagram — circles and perpendicular bisectors

A locus is a set of points that satisfy a given geometric condition. The two most common loci tested in A-Level Mathematics P3 are:

1. Circle: The locus of points $z$ satisfying $|z-a|=r$, where $a$ is a fixed complex number and $r$ is a positive real constant, is a circle with centre at the point representing $a$ and radius $r$.
2. Perpendicular bisector: The locus of points $z$ satisfying $|z-a|=|z-b|$, where $a$ and $b$ are fixed complex numbers, is the perpendicular bisector of the line segment connecting the points representing $a$ and $b$.

To find the Cartesian equation of a locus, substitute $z=x+yi$, expand the modulus expressions, and simplify.

Worked Example

Find the Cartesian equation of the locus $|z+2|=|z-2i|$, and state what type of locus it is.

1. Substitute $z=x+yi$: $|(x+2)+yi|=|x+(y-2)i|$
2. Write modulus expressions: $\sqrt{(x+2{)}^2+y^2}=\sqrt{x^2+(y-2{)}^2}$
3. Square both sides: $(x^2+4x+4+y^2)=x^2+y^2-4y+4$
4. Simplify: $4x=-4y⟹x+y=0$
5. This is the perpendicular bisector of the line segment connecting $(-2,0)$ and $(0,2)$.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Writing the imaginary part including $i$, e.g., $\text{Im}(3-4i)=-4i$. Why it happens: Confusion between the imaginary term and its coefficient. Correct move: The imaginary part is always a real number, equal to the coefficient of $i$, so $\text{Im}(3-4i)=-4$.
• Pitfall 2: Giving arguments in degrees, or outside the $-\pi <\theta \le \pi$ range, e.g., $\arg (-1-i)=\frac{5\pi }{4}$. Why it happens: Forgetting the syllabus specifies principal arguments in radians. Correct move: Always adjust angles by $\pm 2\pi$ to get them in the required range, so $\arg (-1-i)=-\frac{3\pi }{4}$.
• Pitfall 3: Misidentifying the centre of a circle locus, e.g., saying $|z-2+3i|=4$ has centre $(-2,3)$. Why it happens: Misreading the sign inside the modulus. Correct move: Rewrite the expression as $|z-(2-3i)|=4$, so the centre is $(2,-3)$.
• Pitfall 4: Sign errors when squaring modulus terms, e.g., expanding $(x-3{)}^2$ as $x^2-9$. Why it happens: Rushing through algebra to save time. Correct move: Double-check all expansions after squaring, as a single sign error will lead to a completely wrong Cartesian equation and lose you all marks for that part of the question.
• Pitfall 5: Failing to label key features on sketched loci. Why it happens: Focusing on drawing the shape rather than meeting marking criteria. Correct move: Always label the centre and radius for circles, and the two fixed points for perpendicular bisectors, as these are explicit marking points.

8. Practice Questions (A-Level Mathematics P3 Style)

Question 1

(a) Write the complex number $z=-2+2i$ in polar form, giving the principal argument in radians. (b) Hence calculate $z^4$, giving your answer in Cartesian form.

Solution

(a) First calculate the modulus: $|z|=\sqrt{(-2{)}^2+2^2}=\sqrt{8}=2\sqrt{2}$ $z$ lies in the second quadrant, so base angle $\arctan \left(\frac{2}{2}\right)=\frac{\pi }{4}$, principal argument $\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$ Polar form: $z=2\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

(b) Using the polar multiplication rule repeated 4 times: Modulus of $z^4$: $(2\sqrt{2}{)}^4=16\ast 4=64$ Argument of $z^4$: $4\ast \frac{3\pi }{4}=3\pi$, adjusted to principal argument: $3\pi -2\pi =\pi$ So $z^4=64(\cos \pi +i\sin \pi )=64(-1+0i)=-64$

Question 2

(a) Sketch the locus of points $z$ on an Argand diagram satisfying $|z-1-2i|=3$. (b) Find the minimum value of $|z|$ for points on this locus.

Solution

(a) The locus is a circle with centre at $(1,2)$ and radius 3. Sketch the plane, mark the centre point $(1,2)$, draw the circle, and label both the centre and radius.

(b) $|z|$ is the distance from the origin to the point $z$. The minimum distance is the distance from the origin to the centre of the circle minus the radius. Distance from origin to $(1,2)$: $\sqrt{1^2+2^2}=\sqrt{5}\approx 2.236$ Minimum $|z|=3-\sqrt{5}\approx 0.764$

Question 3

Find the Cartesian equation of the locus of points $z$ satisfying $|z-3|=|z+i|$, and state the name of this locus.

Solution

Substitute $z=x+yi$: $$|(x-3)+yi|=|x+(y+1)i|$$ Square both sides: $$(x-3{)}^2+y^2=x^2+(y+1{)}^2$$ Expand and simplify: $$x^2-6x+9+y^2=x^2+y^2+2y+1$$ $$-6x+9=2y+1$$ $$6x+2y=8⟹3x+y=4$$ This is the perpendicular bisector of the line segment connecting the points $(3,0)$ and $(0,-1)$.

9. Quick Reference Cheatsheet

10. What's Next

Complex numbers are a foundational topic for the rest of the A-Level Mathematics P3 syllabus. You will use the content from this guide to solve polynomial equations with complex roots, apply De Moivre's theorem to find powers and roots of complex numbers, and even simplify calculations in later topics like differential equations and vector geometry. Mastering these fundamentals will not only help you score full marks on the dedicated complex number questions (which appear on 90% of past Paper 3 exams) but also reduce your cognitive load when tackling more advanced topics later in the course.

If you have any questions about any part of this guide, or want to practice more exam-style questions tailored to your specific weak spots, head to Ollie, our AI tutor, to get instant explanations, feedback on your working, and custom practice sets. You can also browse our full library of A-Level Mathematics study guides on the homepage to cover all Paper 3 topics ahead of your exam.

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